Find the values in Exercises 9–12.
9. sin(arccos((√2)/2)...

Question

Find the values in Exercises 9–12.

9. sin(arccos((√2)/2))

Accepted Answer

Welcome back, everyone. Evaluate the expression sign of inverse of cosine of -12. A says -12, B 1/2, C negative square root of 3 divided by 2, and D square root of 3 divided by 2. For this problem, let's suppose that. We have an angle theta which is inverse of cosine of -12. So, what we want to do is simply evaluate sine of theta. However, we have to focus on theta itself. We want to understand what type of an angle we have, right? So theta is inverse of cosine of -12, and we have to remember that the principal value of inverts of cosine lies in the interval. Theta belongs to from 0 up to pi inclusive. If we take cosine of both sides, we can show that cosine of theta is equal to -12, right? And it essentially means that cosine is negative, which is true in quadrant number 2. So what we can do is take the positive value of an angle alpha, cosine of alpha would be 12, and we can remember that cosine is equal to 12, when alpha is equal to. I divided by we. And now we can use the reference angle formula. Data is going to be equal to minus alpha, which is minus pi divided by 3, and that's 2 divided by 3. So we now have our angle and we are ready to find sign of that angle. So sin of theta is sign of 2 divided by 3. What we can do now is simply remember that sign of alpha is equal to. Sign Of pi minus alpha, right, we can use that reduction to show that sin of 2 pi divided by 3 is equal to sin of pi minus 2 divided by 3. And that's sign of pi divided by 3, and again we have a perfect angle sine of pi divided by 3 is equal to square root of 3 divided by 2, which corresponds to the answer choice d. Thank you for watching.

Find the values in Exercises 9–12.9. sin(arccos((√2)/2))

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Find the values in Exercises 9–12.
9. sin(arccos((√2)/2))